How many milliliters of 0.0459 MH3PO4 could be completely neutralized by 46.2 mL of 0.136 M KOH? The balanced equation for the reaction is H3PO4(aq) + 3KOH(aq)→ K3PO4(aq) +3H20 V = mL
How many milliliters of 0.0459 MH3PO4 could be completely neutralized by 46.2 mL of 0.136 M KOH? The balanced equation for the reaction is H3PO4(aq) + 3KOH(aq)→ K3PO4(aq) +3H20 V = mL
Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter3: Molecules, Moles, And Chemical Equations
Section: Chapter Questions
Problem 3.85PAE: 3.85 The particulate drawing shown represents an aqueous so- lution of an acid HA, where A might...
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How many milliliters of 0.0459 M H3PO4 Could be completely neutralized by 46.2 mL of 0.136 M KOH? The balanced equation for the
reaction is
H3PO4(aq) + 3KOH(aq)→K3PO4(aq) + 3H,O
V =
mL
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How many milliliters of 0.0459 M H3PO4 Could be completely neutralized by 46.2 mL of 0.136 M KOH? The balanced equation for the
reaction is
H3PO4(aq) + 3KOH(aq)→K3PO4(aq) + 3H,O
V =
mL
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A ô . E O A )
144
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8.
[C
E
D
K
pause
C
ctri
alt
T.
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